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How many different three letter permutations can be from the letter in the word clipboard?
Robmedown7
There are 9 * 8 * 7, or 504, three letter permutations that can be made from the letters in the work CLIPBOARD.
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∙ 12y ago
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How many different three letter permutations can be formed from the letters in the word clipboard?
There are three that I can see, there's clip, board and lip.
How many different 4 letter permutations are there for the letters in the word toolroom?
120 four letter permutations if you don't allow more than one 'o' in the four letterarrangement.209 four letter permutations if you allow two, three and all four 'o'.1.- Let set A = {t,l,r,m,}, and set B = {o,o,o,o}.2.- From set A, the number of 4 letter permutations is 4P4 = 24.3.- 3 letters from set A give 4P3 = 24, and one 'o' can take 4 different positions in theword. That gives us 24x4 = 96 four letter permutations.4.- In total, 24 + 96 = 120 different four letter permutations.5.- If the other three 'o' are allowed to play, then you have 2 letters from set A thatgive 4P2 = 12 permutations and two 'o' can take 4C2 = 6 position's, giving 12x6 = 72four letter permutations.6.- One letter from set A we have 4P1 = 4, each one can take 4 different positions, therest of the spaces taken by three 'o' gives 4x4 = 16 different permutations.7.- The four 'o' make only one permutation.8.- So now we get 72 + 16 + 1 = 89 more arrangements adding to a total of 89 + 120 = 209 different 4 letter arrangements made from the letters of the word toolroom.[ nCr = n!/((n-r)!∙r!); nPr = n!/(n-r)! ]
How many permutations for 112?
Three.
What are the number of permutations of the letters of the word MATH?
For the first letter, you can choose any of the four. For the second letter, you can choose any of the remaining three. For the third letter, you can choose either of the remaining two. For the fourth letter, you don't get a choice, you have to use the one that's left. So 4x3x2x1 = 24 possible permutations.
How many distinguishable permutations of letters are in the word queue?
three
How many different three letter permutations can be formed from the letters in the word clipboard?
There are three that I can see, there's clip, board and lip.
How many different 4 letter permutations are there for the letters in the word toolroom?
120 four letter permutations if you don't allow more than one 'o' in the four letterarrangement.209 four letter permutations if you allow two, three and all four 'o'.1.- Let set A = {t,l,r,m,}, and set B = {o,o,o,o}.2.- From set A, the number of 4 letter permutations is 4P4 = 24.3.- 3 letters from set A give 4P3 = 24, and one 'o' can take 4 different positions in theword. That gives us 24x4 = 96 four letter permutations.4.- In total, 24 + 96 = 120 different four letter permutations.5.- If the other three 'o' are allowed to play, then you have 2 letters from set A thatgive 4P2 = 12 permutations and two 'o' can take 4C2 = 6 position's, giving 12x6 = 72four letter permutations.6.- One letter from set A we have 4P1 = 4, each one can take 4 different positions, therest of the spaces taken by three 'o' gives 4x4 = 16 different permutations.7.- The four 'o' make only one permutation.8.- So now we get 72 + 16 + 1 = 89 more arrangements adding to a total of 89 + 120 = 209 different 4 letter arrangements made from the letters of the word toolroom.[ nCr = n!/((n-r)!∙r!); nPr = n!/(n-r)! ]
How many different numbers can be formed with the digits 345?
Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.
How many permutations for 112?
Three.
What are the number of permutations of the letters of the word MATH?
For the first letter, you can choose any of the four. For the second letter, you can choose any of the remaining three. For the third letter, you can choose either of the remaining two. For the fourth letter, you don't get a choice, you have to use the one that's left. So 4x3x2x1 = 24 possible permutations.
Why do humans see different colors?
You have receptors for three primary colours in your retina. You can perceive the thousands of colours due to stimulation of the these receptors at different intensity in various permutations and combinations.
How many distinguishable permutations of letters are in the word queue?
three
How many license plates can be made using three digits and three letters without repetition and alternating letters and digits?
If it alternates and you start with a letter, then there are 11,232,000 permutations. Then if you start with a number and alternate, there are another 11,232,000 permutations, for a total of 22,464,000 permutations. If you exclude the letters I, S, B, and O (because they look kind-of like 1, 5, 8, & 0 - kind-of important on license plates) then you are down to 6,652,800 & 13,305,600 respectively.
How many permutations are there in a 3-letter code which can use the letters from A to Z inclusive?
A - Z means you can use the whole alphabet, which usually contains 26 letters. So a one-letter code would give you 26 permutations. 2 letters will give you 26 x 26 permutations. A three letter code, finally, will give you 26 x 26 x 26 , provided you don't have any restrictions given, like avoiding codes formed from 3 similar letters and such.
What number of permutation of the letters in the word smart?
For the first letter, you can have any of 5 choices. Then for the next, you can have any of four. Then for the next, three and so on. Thus the number of permutations can be calculated by 5x4x3x2x1. Doing this gives 120. Therefore the number of permutations of the letters in the word smart is 120.
How many distinguishable permutations are there of the letters in the word effective?
The number of permutations of the letters EFFECTIVE is 9 factorial or 362,880. To determine the distinct permutations, you have to compensate for the three E's (divide by 4) and the two F's (divide by 2), giving you 45,360.
How many permutations are possible of the word yard?
Imagine you have four empty buckets in which to put any of the four letters Y, A, R, D into. At first you have four letters that can be placed in any of the four empty buckets. Once you've placed a letter in a bucket you only have three letters and three empty buckets to choose from. And so on... So there are 4x3x2x1 = 24 permutations of the word YARD.
